Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 4e5 + 10;
int h[N], ne[N], e[N], idx;
int dfn[N], ins[N], low[N], times;
int depth[N], fa[N], id[N], top[N], son[N], sz[N];
int n, m, q;
vector<int> num[N];
struct arr {
  int q[10], len;
  int size() { return len; }
  void push_back(int x) { q[len++] = x; }
  void clear() {
    for (int i = 0; i < 10; i++) {
      q[i] = 0;
    }
    len = 0;
  }
};
struct node {
  int l, r;
  arr s;
} tr[N << 2];
void add(int a, int b) { e[idx] = b, ne[idx] = h[a], h[a] = idx++; }
void dfs(int u, int f) {
  sz[u] = 1;
  int i;
  for (i = h[u]; i != -1; i = ne[i]) {
    int j = e[i];
    if (j == f) continue;
    fa[j] = u;
    depth[j] = depth[u] + 1;
    dfs(j, u);
    sz[u] += sz[j];
    if (sz[j] > sz[son[u]]) son[u] = j;
  }
}
void dfs1(int u, int x) {
  dfn[u] = ++times;
  id[times] = u;
  top[u] = x;
  if (!son[u]) return;
  dfs1(son[u], x);
  int i;
  for (i = h[u]; i != -1; i = ne[i]) {
    int j = e[i];
    if (j == fa[u] || j == son[u]) continue;
    dfs1(j, j);
  }
}
arr merge(arr a, arr b) {
  arr d;
  d.clear();
  int i = 0, j = 0;
  while (j < b.size() && i < a.size() && d.size() < 10) {
    if (a.q[i] < b.q[j])
      d.push_back(a.q[i++]);
    else
      d.push_back(b.q[j++]);
  }
  while (j < b.size() && d.size() < 10) d.push_back(b.q[j++]);
  while (i < a.size() && d.size() < 10) d.push_back(a.q[i++]);
  return d;
}
void build(int u, int l, int r) {
  if (l == r) {
    tr[u] = {l, r};
    int i;
    for (auto x : num[l]) {
      tr[u].s.push_back(x);
      if ((int)tr[u].s.size() == 10) break;
    }
  } else {
    tr[u] = {l, r};
    int mid = l + r >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
    tr[u].s = merge(tr[u << 1].s, tr[u << 1 | 1].s);
  }
}
arr q1;
void query(int u, int l, int r) {
  if (tr[u].l >= l && tr[u].r <= r) {
    q1 = merge(q1, tr[u].s);
    return;
  }
  int mid = tr[u].l + tr[u].r >> 1;
  if (l <= mid) {
    query(u << 1, l, r);
  }
  if (r > mid) query(u << 1 | 1, l, r);
}
void querypath(int x, int y, int z) {
  q1.clear();
  while (top[x] != top[y]) {
    if (depth[top[x]] < depth[top[y]]) swap(x, y);
    query(1, dfn[top[x]], dfn[x]);
    x = fa[top[x]];
  }
  if (depth[x] > depth[y]) swap(x, y);
  query(1, dfn[x], dfn[y]);
  int k = min((int)q1.size(), z);
  printf("%d ", k);
  for (int i = 0; i < k; i++) {
    printf("%d ", q1.q[i]);
  }
  printf("\n");
}
int main() {
  memset(h, -1, sizeof h);
  cin >> n >> m >> q;
  int i;
  for (i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    add(a, b);
    add(b, a);
  }
  depth[1] = 1;
  dfs(1, -1);
  dfs1(1, 1);
  for (i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    num[dfn[x]].push_back(i);
  }
  build(1, 1, n);
  while (q--) {
    int u, v, a;
    scanf("%d%d%d", &u, &v, &a);
    querypath(u, v, a);
  }
}